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User blog:Ynought/A new function idea?
ATTENTION:THIS IS ILLDEFINED FROM BEGINNING TO END!! What is this about? I hope to make this the next step after hydras,which i call Active Head shift Eh,this is really ill-defined as of now;This "definition" is simply a general outline Rules and procesees Each node in the tree?hydra?(i will call it monster from now) has to be labeled with the smallest number this and keep increasing the number at the bottom of the \(V\) and then continue until you have named all nodes two nodes cannot overlap and but all of them have the same size \(f(V_i^j)\) is the value assigned too \(V_i^j\) \(M_1\) is the starting monster \(M_{i+1}\) is the monster that is the result after aplying the rules too \(M_i\) \(M_j\) consists of nodes on a plane where every \(V_k^j\) has to be connected to at least one other \(V_i^j\) So now the hard part comes:The ruleset: The active node will be found like this: 1.let \(l\) be the number off nodes: 2.if \(l=1\): 2.1. then \(M_1^j\) is the active node 3 if \(l=2\): 3.1.\(f(V_1^j)>f(V_2^j)\) then \(V_2\) is the active node 3.2..\(f(V_1^j)2\) then: 4.1. check for the largest \(f(V_i^j)\) 4.1.1 if it is the root node then use that 4.1.2 if it is not the root node then take the node with the largest \(f(V_i^j)\) which is connected to the node from 4.1. and make that the root node Here comes the hard part the solving:I most likely completely failed. But here it is:(W.i.p.) 1.let: let \(G(a,b)\) be defined as the graph of an regular \(a\)-agon,with nodes at the corners conected via lines, that is a smallest valid monster that cant fit inside a square of \(b\times b\) units \(\{N\Rightarrow x\}\) be the set of nodes connected via a line in the graph to node x(x has to be a defined node) \(\{N\Leftarrow\mathbb{x}\}\) be solved by: #check if it is possible to place a node (with its center) four units above the Center of \(N\) #1. if yes then place one there and conect them via a line #2. if no then repeat from step one but one unit to the tight #3. if after repeating 1.1 and 1.2 indefinetly and it is never possible then move the vertex you want to place one up Repeat the processs above \(\text{cardinality}(\mathbb{x})\) times,all vertices from the process above have the \(f\) value of the \(N\) \(N_A\) is the active node now let \(L(G(a,b))\) be solved by: 2.Apply \(\{N_A\Leftarrow\{N\Rightarrow N_A \}\}\) 3.Draw all possible Lines (That dont overlap with a vertex) and let \(j=\sum_0 ^{k=n|n=\text{max}(k|\exists V_k)}f(V_k)\) 4. replace the active node with a copy of the current monster 5. delete the current active node and add 1 to all values of the nodes that are at least 2 smaller than the deleted active node,decrease j by 1 (Also in the copy,if there is a \(f(V_k\) bigger than that of the avtive node,then decrease the value of \(V_k\) to be as big as that of \(N_A\) 6. if \(j\neq0\) go to step 4 otherwise continue 7. if there is only one node left; 7.1 go to 2. 7.2.otherwise if the largest \(f(V_i)>|\exists V_i|\); 7.2.1 add all \(f(V_i)\) still left and call that \(L(GM)\) 7.2.2 otherwise add 1 to every \(f(V_i)\) still left and go to 2. 8.Let \(L(G(a,b))=L(GM)\) for \(G(a,b)\) Value of M Finally let \(M(a,b)=L(G(a,b))\) Now i have now idea if this terminates,and if it does then i have no idea how fast it grows Category:Blog posts